In a galvanic cell the cathode is an $\mathrm{Ag}^{+}(1.00 \mathrm{M}) / \mathrm{Ag}(s)$ half-cell. The anode is a standard hydrogen electrode immersed in a buffer solution containing \(0.10 \mathrm{M}\) benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) and $0.050 \mathrm{M}\( sodium benzoate \)\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-} \mathrm{Na}^{+}\right)\(. The measured cell voltage is \)1.030 \mathrm{~V}\(. What is the \)\mathrm{p} K_{\mathrm{a}}$ of benzoic acid?

The Nernst equation is used to calculate the potential of a half-cell at a given concentration. The equation is: \[ E_{cell} = E^\circ - \frac{0.05916}{n} \log Q \] where \(E_{cell}\) is the potential, \(E^\circ\) is the standard potential, \(n\) is the number of electrons transferred in the redox reaction, and \(Q\) is the reaction quotient. For the Ag+/Ag half-cell, the standard potential \(E^\circ = 0.7996\ V\), and the number of electrons transferred \(n = 1\), since the reduction reaction is: \[ \mathrm{Ag}^{+} + 1e^{-} \rightarrow \mathrm{Ag}(s) \] The reaction quotient, \(Q\), for the reduction reaction is: \[ Q = \frac{[\mathrm{Ag}^{+}]}{[\mathrm{Ag}(s)]} \] Since the activity of a solid is defined as 1, we can rewrite the reaction quotient as: \[ Q = [\mathrm{Ag}^{+}] \] We are given that the concentration of Ag+ ions is 1.00 M. Plugging the values into the Nernst equation, we get: \[ E_{\mathrm{Ag}^{+}/\mathrm{Ag}} = 0.7996 - \frac{0.05916}{1} \log(1) = 0.7996\ V \]

Since we know the overall cell voltage (\(1.030\ V\)), we can use the cell voltage equation: \[ E_{cell} = E_{cathode} - E_{anode} \] Plugging in the values we have found, we get: \[ 1.030 = 0.7996 - E_{anode} \] Solving for the anode potential, we get: \[ E_{anode} = 0.7996 - 1.030 = -0.2304\ V \]

Now, we need to relate the potential of the hydrogen half-cell to the pH of the solution. For the hydrogen half-cell, the standard potential \(E^\circ = 0\ V\), and the number of electrons transferred \(n = 2\), since the reduction half-reaction is: \[ 2\mathrm{H}^{+} + 2e^{-} \rightarrow \mathrm{H}_{2}(g) \] Under standard conditions, \(\frac{[\mathrm{H}^{+}]}{P_{\mathrm{H}_{2}}}\) should be 1. However, we need to consider the non-standard conditions which can be done using the following formula at \(25\degree\)C: \[ E_{anode} = - \frac{0.05916}{2} \log([\mathrm{H}^{+}]) \] We have calculated the anode potential, so we can solve for the concentration of H+ ions: \[ -0.2304 = - \frac{0.05916}{2} \log([\mathrm{H}^{+}]) \] This gives us: \[ \log([\mathrm{H}^{+}]) = 7.78 \] and now we can calculate pH of the cell: \[ \mathrm{pH} = -\log([\mathrm{H}^{+}]) = -7.78 \]

Using the Henderson-Hasselbalch equation, we can find the pKa of benzoic acid by plugging in the concentrations of the benzoate and the benzoic acid: \[ \mathrm{pH} = \mathrm{p} K_{a} + \log \frac{[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{COO}^{-}]}{[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{COOH}]} \] Substitute the pH value and concentrations of benzoate and benzoic acid in the equation: \[ -7.78 = \mathrm{p} K_{a} + \log \frac{0.050}{0.10} \] Now we can solve for the pKa of benzoic acid: \[ \mathrm{p} K_{a} = -7.78 - \log \frac{1}{2} \approx - 7.48 \] Therefore, the pKa of benzoic acid in this galvanic cell is approximately -7.48.